A biconvex lens of refractive index mu -1- - is immersed in a liquid of refractive index mu -2- - If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice ismu -1-mu -2-mu -1-mu -2-mu -1-mu -2-mu -2-dfrac-mu -1-2-

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Toppr Admin · Accepted Answer

-x3BC-1-x2192-x3BC-21f-x3BC-eoq-x2212-1-1R1-x2212-1R2-divergent-x21D2-x3BC-1-x3BC-2-x2212-1-lt-0-x21D2-x3BC-1-lt-x3BC-2since biconvex-x21D2-1R1-x2212-1R2-gt-0

A biconvex lens of refractive index $μ_{1}$ , is immersed in a liquid of refractive index $μ_{2}$ . If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice is
$μ_{1} > μ_{2}$
$μ_{1} < μ_{2}$
$μ_{1} = μ_{2}$
$μ_{2} = \frac{μ_{1}}{2}$

$μ_{1} \to μ_{2}$

$\frac{1}{f} = (μ_{e o q} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$d i v e r g e n t \Rightarrow \frac{μ_{1}}{μ_{2}} - 1 < 0 \Rightarrow μ_{1} < μ_{2}$

since $b i c o n v e x \Rightarrow (\frac{1}{R_{1}} - \frac{1}{R_{2}}) > 0$

A biconvex lens of refractive index μ1 , is immersed in a liquid of refractive index μ2 . If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice isμ1>μ2μ1<μ2μ1=μ2μ2=μ12

μ1→μ21f=(μeoq−1)(1R1−1R2)divergent⇒μ1μ2−1<0⇒μ1<μ2since biconvex⇒(1R1−1R2)>0

A biconvex lens of refractive index $μ_{1}$ , is immersed in a liquid of refractive index $μ_{2}$ . If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice is
$μ_{1} > μ_{2}$
$μ_{1} < μ_{2}$
$μ_{1} = μ_{2}$
$μ_{2} = \frac{μ_{1}}{2}$

$μ_{1} \to μ_{2}$

$\frac{1}{f} = (μ_{e o q} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$d i v e r g e n t \Rightarrow \frac{μ_{1}}{μ_{2}} - 1 < 0 \Rightarrow μ_{1} < μ_{2}$

since $b i c o n v e x \Rightarrow (\frac{1}{R_{1}} - \frac{1}{R_{2}}) > 0$